Question

A voltaic cell is constructed with two silver-silver chloride electrodes, where the halfreaction is AgCl (s) + e- → Ag (s) + Cl- (aq) E° = +0.222 V
The concentrations of chloride ion in the two compartments are 0.0222 M and 2.22 M, respectively. The cell emf is __________ V.

A) 0.212

B) 0.118

C) 0.00222

D) 22.2

E) 0.232

Answer 1

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  $Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-$

Reduction half reaction (cathode):  $AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}]_{diluted}$ = 0.0222 M

$[Cl^{-}]_{concentrated}$ = 2.22 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Hence, the cell potential of the cell is +0.118 V